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3 years ago

13

What is extraneous roots?

1 answer:

kobusy [5.1K]3 years ago

7 0

extraneous root is a solution o an equation that seems to be right but when we check it (by substituting it into the original equation) it turns out not to be right

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Find the mean and median of the following data set:

VARVARA [1.3K]

Answer:

Mean: 23. 54545455

Median: 22

Step-by-step explanation:

The median represents the center the most because when you are solving for the median, you look for the direct center of the terms which will be 22, the median.

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2 years ago

The cook has a variety of meals to prepare for a villa center’s residents. she averages 16 vegetarian meals every day of the wee

MrRissso [65]

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3 years ago

Find the simple interest for :<br> $668,5%, 2 years

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2 years ago

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What is V36x4 in simplest form?

DochEvi [55]

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3 years ago

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Evaluate the integral by making an appropriate change of variables.

VARVARA [1.3K]

By inspecting the integrand, the "obvious" choice for substitution would be

<em>u</em> = <em>y</em> + <em>x</em>

<em>v</em> = <em>y</em> - <em>x</em>

<em />

Solving for <em>x</em> and <em>y</em>, we would have

<em>x</em> = (<em>u</em> - <em>v</em>)/2

<em>y</em> = (<em>u</em> + <em>v</em>)/2

in which case the Jacobian and its determinant are

$J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12$

The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.

Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So

-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>

-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2

-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>

-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>

<em />

So, the integral becomes

$\displaystyle\iint_R5\cos\left(7\frac{y-x}{y+x}\right)\,\mathrm dA=\int_8^9\int_{-u}^u\frac52\cos\left(\frac{7v}u\right)\,\mathrm dv\,\mathrm du$

$=\displaystyle\frac52\int_8^9\frac u7(\sin7-\sin(-7))\,\mathrm du$

$=\displaystyle\frac57\sin7\int_8^9u\,\mathrm du$

$=\displaystyle\frac5{14}\sin7(9^2-8^2)=\boxed{\frac{85}{14}\sin7}$

4 0

2 years ago